simple matrix

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Austin
Austin 2011-4-20
how can i write the code that makes a 10X10 matrix that looks like this
00000000000
00000000001
00000000011
00000000111
00000001111
00000011111
00000111111
00001111111
00011111111
00111111111
01111111111

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回答（3 个）

Oleg Komarov
Oleg Komarov 2011-4-20
fliplr(tril(ones(10),-1))
or
rot90(tril(ones(10),-1))
EDIT#2 (per Matt's suggestion)
Timings of:
a) fliplr
b) rot90
c) bsxfun
function M = fliptril(N,n)
% Obligatory timings....
times = zeros(N,3);
for ii = 1:N
tic
a = fliplr(tril(true(n),-1));
times(ii,1) = toc;
tic
b = rot90(tril(true(n),-1));
times(ii,2) = toc;
tic
c = bsxfun(@lt,(n:-1:1).',1:n);
times(ii,3) = toc;
end
m = mean(times(2:end,:));
M = m/min(m);
Timings N = 10,000, n = 100 (fliplr is faster):
>> M = fliptril(10000,100)
M =
1.0000 1.3644 2.2590
Timings N = 1,000, n = 500 (bsxfun is faster):
>> M = fliptril(1000,500)
M =
1.2637 1.7527 1.0000
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Oleg Komarov
Oleg Komarov 2011-4-21
WinVista 32, r2010b.
On http://www.mathworks.com/products/matlab/whatsnew.html r2009b bsxfun supports multithreading
Walter Roberson
Walter Roberson 2011-4-21
I thought that perhaps calculating dt=1:n first and using flipud(dt(:)) and dt as the bsxfun arguments might speed things up, but on my tests they slow things down.
In 2008b, I find that the bsxfun approach is the slowest, and that the ratio gets worse as n increases.
Ah, correction: calculating dt and using flipud(dt(:)) is a bit faster than the original method once n passes about 2000. And by 4000, the bsxfun() methods are faster than rot90, with the ratio getting worse for rot90 as n increases.
Here are the ratios for n = 50000 (last column is with calculating 1:n and flipud() it)
1 1.74457576765052 1.1147589050359 1.11381126664567
Here are the ratios for n = 1000
1 1.27935606060606 1.82795055821372 1.83313397129187
thus rot90 is efficient for smaller n and bsxfun is not, but for larger n, rot90 and bsxfun have pretty much swapped efficiency ratios.

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Walter Roberson
Walter Roberson 2011-4-20
Look Ma, no rotate!
bsxfun(@lt,(10:-1:1).',1:10)
Jan
Jan 2011-4-26
For larger matrices BUFFER is faster than FLIPLR(TRIL)) under Matlab 2009a:
n = 50;
data = ones(1, 2*n-1);
data(1:50) = 0;
M = buffer(data, n, n-1, 'nodelay');

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